📝 题目
例 3 设 $m \in \mathbb{N}$ ,试求函数 $f\left( x\right) = {x}^{-m}\left( {x \neq 0}\right)$ 的导数.
💡 答案与解析
解 我们有
$$ \frac{f\left( {x + h}\right) - f\left( x\right) }{h} = \frac{1}{h}\left\lbrack {\frac{1}{{\left( x + h\right) }^{m}} - \frac{1}{{x}^{m}}}\right\rbrack $$
$$ = \frac{1}{h}\left( {\frac{1}{x + h} - \frac{1}{x}}\right) \left\lbrack {\frac{1}{{\left( x + h\right) }^{m - 1}} + \frac{1}{{\left( x + h\right) }^{m - 2}x} + \cdots + \frac{1}{{x}^{m - 1}}}\right\rbrack $$
$$ = - \frac{1}{\left( {x + h}\right) x}\left\lbrack {\frac{1}{{\left( x + h\right) }^{m - 1}} + \frac{1}{{\left( x + h\right) }^{m - 2}x} + \cdots + \frac{1}{{x}^{m - 1}}}\right\rbrack , $$
因而
$$ {f}^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {x + h}\right) - f\left( x\right) }{h} $$
$$ = - \frac{m}{{x}^{m + 1}} = - m{x}^{-m - 1}. $$